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3 years ago

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Julia has 2 identical rooms in her house. if each room measures 8 feet on one side and 12 feet on another, what is the total are

a of the rooms

1 answer:

Jlenok [28]3 years ago

5 0

First we have the work out the area of 1 room. So that is 8ft * 12ft which is 96ft^2. And if there are 2 of these rooms, then we multiply this value by 2. Giving us a total area of 192ft^2.

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You need to cross a canal and want to determine the distance across the opposite side. Since you're able to take measurements on

MariettaO [177]

Answer:

36.7 ft

Step-by-step explanation:

Measurements for the sides of the canal is given as: 40 ft. and 16 ft.

You solve the above question using Pythagoras Theorem

The distance across the canal is calculated as:

√(40² - 16²)

= √(1344)

= 36.66060556 ft

Approximately = 36.7 ft

Therefore, the distance x across the canal = 36.7 ft

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3 years ago

Malik is making 36 small pack of chili spices and 15 large packs of chili spices. A small pack contains 2.35 ounces of spices.A

Elena L [17]

The answer is
To find how many ounces you need to make, you multiply the number of packs by how many ounces.
So, it would look like:
36*2.35=84.6
15*10.8=162

Then you add the ounces together and you have your answer:
162+84.6=246.6 ounces

5 0

3 years ago

Read 2 more answers

How many tens are in 120,000

Marina CMI [18]

12000
Just do
120000 divided by 10

8 0

3 years ago

Read 2 more answers

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

7m - 17 = -24<br> what is m?

aev [14]

Answer:

Move all terms that don't contain

m

to the right side and solve.

Exact Form:

m

=

41

7

Decimal Form:

m

=

5.

¯¯¯¯¯¯¯¯¯¯¯¯

857142

Mixed Number Form:

m

=

5

6

7

Tap to view steps...

3 0

3 years ago